Let S=left{Z in C: bar{z}=ileft(z^2+operatorn | vedclass

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Question

Let $Z=x+$ iy, $x \in R, y \in R$

$x-i y=i\left(x^2-y^2+(2 x y) i+x\right)$

$x =- 2 x x$

$- y =- y ^2+ x ^2+ x$

$\Rightarrow x=0, y=-\frac

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$4$

Vedclass · Answer

Let $Z=x+$ iy, $x \in R, y \in R$

$x-i y=i\left(x^2-y^2+(2 x y) i+xight)$

$x =- 2 x x$

$- y =- y ^2+ x ^2+ x$

$\Rightarrow x=0, y=-\frac{1}{2}(	ext { from }(1))$

If $x 
eq 0$, then $y =0,1$

If $y =-\frac{1}{2}$, then $x =\frac{1}{2},-\frac{3}{2}$

$Z =0+ i 0,0+ i , \frac{1}{2}-\frac{ i }{2},-\frac{3}{2}-\frac{ i }{2}$

Let $S=\left\{Z \in C: \bar{z}=i\left(z^2+\operatorname{Re}(\bar{z})\right)\right\}$. Then $\sum_{z \in S}|z|^2$ is equal to

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